Calculating force on a string and harmonics, something doesnt add up

I had to calculate the force on a wire and its second harmonic. Given data were, the diameter of of the string 1,2 mm , length of 1,25m. Density of steel 7,8 kg/dm^3, its fundamental frequency is 300 Hz, and also one end is fixed the other is loose so it can move frictionless up and down.

I know of a very handy equation but it should work when both ends are fixed.

may I use this equation if the other end is lose? What is the difference when having the other end loose. I never paid much attention to that, I practiced only on both ends fixed, since that makes sense to me. Why would anybody want to have the other end loose?

anyway, I did some algebra on the equation and found that the force should be around 6 kN! I couldnt believe my eyes. Surely I have misscalculated. But oddly, all the people I talked to get even higher value! Most got around 20kN but that is impossible! Imagine having 2 tone weight to put the force needed on a string with diameter 1,2 mm!

Since I am not sure if my reasoning is correct but a steel wire should surely break at such high values! If I recall correctly from reading a text. Breaking Force = Breaking Stress * Area. If I recall Breaking Stress for piano steel wire is at most 2000 MPa or lets say 2 GPa. But that would be an overestimate but never mind. Putting this into the equation I get Breaking Force = 2 * 10^9 * Pi * (0,6 * 10^-3 m)^2= 2,2kN. Wow. But the problem said it was common steel wire soo that is half the value!

Something is wrong here, either my calculations/reasoning, or the problem itself make no sense. Also the people I talked to got the second harmonic to be 900 Hz. Is it just me or is that really really (too) high value?

Did you:
convert the diameter to radius (in meters)?
Did you convert the density to kg/m^3?

I don't know how a string vibrates under tension with a "loose end." The tension would have to come from its own weight, and thus the fundamental would change with the height of the string, and still no way could the required tension be provided.

If it were to oscillate with one "open" end, it would act according to the pipe open at one end formula: f_n=nv/4L BUT
with an open-end oscillator, there are no "even harmonics." n=1, 3, 5, 7, 9...

I can imagine the fenomena a bit better now. I imagine to have 2 poles that are vertical, and the string is in horizontal between then, fixed to a pole on one end and having a loop like thing on the other pole to enable it to move up and down. Soo the force of tension would vary depending on the up or down displacement of the other end. Gravity should be turned off...I imagine the poles to be horizontal with the ground.

yeah, the correct equation would be the

f_n=nv/4L

i presume I can use v = sqrt(F/mass per unit length)

soo f_1=sqrt(F/micro)/4*length, some algebra and I have

F = mass per unit lengt * 4^2 * length^2*freq^2 to solve for the data of the problem I have
F = 18,3 kN!!!

I get the ball park of what most people got on the exam.

The second question was to calculate the next higher harmonic not the second, my mistake. I was to quick to write. It is logical then to be 3 times the first harmonic or 900 Hz.

It is clear now. I made the mistake of not drawing a picture of the fenomena, I should have known better, always draw the picture no matter how silly it may be. It is bad problem anyway since you don't have a real phenomena behind it, I had to invent one to picture it.

Soo the answers are correct, but how can you have 18,3 kN force on the string? The problem might now be mathematically correct but is it possible to have 18,3 kN force on a steel string? Am I correct to asume that at most a force of 2 kN can be applied to a steel string before it breaks?